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# Exercise 3.4

7th Maths : Term 2 Unit 3 : Algebra : Exercise 3.4 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

Exercise 3.4

Miscellaneous Practice problems

1. 62× 6m = 65, find the value of ‘m’.

62+m = 65

Base are equal. So powers also equal.

2 + m =5

m = 5–2 = 3

m = 3

2. Find the unit digit of 124128 × 126124 .

Unit digit of base 124 is 4 and power is 128 (Even number).

So, the unit digit of 124128 is 6.

Unit digit of base 126 is 6 and power is 124 (Even number).

So, the unit digit of 126124 is 6.

6 × 6 = 36, Unit digit of 36 is 6.

So, Unit digit of 124128 × 126124 is 6.

3. Find the unit digit of the numeric expression: 1623 + 7148 + 5961

Unit digit of base 16 is 6 and power is 23 (Odd number).

So, the unit digit of 1623 is 6.

Unit digit of base 71 is 1 and power is 48 (Even number).

So, the unit digit of 7148 is 1.

Unit digit ofbase 59 is 9 and power is 61 (Odd number).

So, the unit digit of 5961 is 9.

Adding Unit digits, 6+1+9 = 16, Unit digit of 16 is 6.

So, unit digit of 1623 + 7148 + 5961 is 6.

4. Find the value of . [ ( 1)6 × ( 1)7 × ( 1)8 ] / [ ( 1)3 × ( 1)5 ]

[ (–1)6 × (–1)7 × (–1)8 ] / [ (–1)3 × (–1)5 ]

(–1)6 = 1, (–1)–1 = –1, (–1)8 = l, (–1)3 = –1, (–1)5 = –1

[ (–1)6 × (–1)7 × (–1)8  ] / [ (–1)3 × (–1)5 ] = [ 1 × (–1) ] / [ (–1) × (–1) ] = (–1) / 1 = –1

5. Identify the degree of the expression, 2a 3bc + 3a3b + 3a3c 2a2b 2 c2

Degree of 2a3bc is 5

Degree of 3a3b is 4

Degree of 3a3c is 4

Degree of 2a2b2c2 is 6

The degree of the expression is 6.

6. If p = −2, q = 1 and r = 3 , find the value of 3p2q2r .

p = –2, q = l, r = 3

3p2q2r = 3(–2)2 (1)2 × 3

= 3 × 4 × 1 × 3 = 36

Challenge Problems

7. LEADERS is a WhatsApp group with 256 members. Every one of its member is an admin for their own WhatsApp group with 256 distinct members. When a message is posted in LEADERS and everybody forwards the same to their own group, then how many members in total will receive that message?

Number of WhatsApp group members = 256

Number of members in each group = 256

Number of members in total will receive that message = 256 × 256

Number of members in total will receive that message = 65536

8. Find x such that 3x +2 = 3x + 216 .

3x + 32 –3x = 216

3x + 9 –3x = 216

3x (9–1) = 216

3x  × 8 = 216

3x = 216 / 8 = 27 = 33

3x = 33

Bases are equal.

So, the powers also equal.

x = 3.

9. If X = 5x 2 + 7x + 8 and Y = 4x 2 7x + 3 , then find the degree of X+Y.

X = 5x2 + 7x + 8

Y = 4x2 – 7x + 3

X + Y = 9 x2 +11

The degree of X+Y is 2.

10. Find the degree of ( 2a 2 + 3ab b2 ) ( 3a 2 ab 3b2 )

2a2 + 3ab –b2 – 3a2 + ab + 3b2

2a2 – 3a2 + 3ab + ab – b2 + 3b2

–a2 + 4ab + 2b2

The degree of the expression is 2.

11. Find the value of w, given that x = 3 , y = 4 , z = −2 and w = x2 y 2 + z2 xyz .

w = x2 –y2 + z2 –xyz

w = (3)2 – (4)2 + (–2)2 – 3 × 4 × –2

= 9 – 16 + 4 + 24

= 9 + 4 + 24 – 16

= 37 – 16 = 21

12. Simplify and find the degree of 6x 2 + 1 [ 8x {3x2 7 ( 4x2 2x + 5x + 9)}]

6x2 + 1 – [8x– {3x2 –7 – (4x2 – 2x + 5x + 9)}]

6x2 + 1 – [8x –3x2 + 7 + (4x2 –2x + 5x + 9)]

6x2 + 1 – 8x + 3x2 –7 – (4x2 + 2x –5x –9)

6x2 + 3x2 – 4x2 –8x – 5x + 2x+ 1–7–9

5x2 – 13x + 2x + 1 –16

5x2 –11x –15

The degree of the expression is 2.

13. The two adjacent sides of a rectangle are 2x2 5xy + 3z2 and 4xy x 2 z2 . Find the perimeter and the degree of the expression.

Length l = 2x2 –5xy + 3z2

Breadth b = 4xy –x2 –z2

Perimeter = 2(l + b)

= 2(2x2 – 5xy + 3z2 + 4xy – x2 – z2)

= 2(x2 – xy + 2z2)

Perimeter of the rectangle = 2x2 – 2xy + 4z2

The degree of the Perimeter is 2.

Exercise 3.4

1. m = 3

2. 6

3. 6

4. −1

5. 6

6. 36

Challenge problems

7. 65536

8. x = 3

9. 2

10. 2

11. 21

12. 5 x2 − 11x − 5 ; 2

13. 2x2 − 2xy + 4z2 ; 2

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7th Maths : Term 2 Unit 3 : Algebra : Exercise 3.4 | Questions with Answers, Solution | Algebra | Term 2 Chapter 3 | 7th Maths